Download Basic Principles of Heredity Notes AP Biology Mrs. Laux

Basic Principles of Heredity Notes AP Biology Mrs. Laux
I. First ideas of heredity
-blending inheritance
1. children received traits from parents-result: blend of 2 parents
2. according to this:
a. uniform appearance after a few generations
b. blended genes cannot be separated
3. Problems:
a. no uniform appearance (variation)
b. some traits hide and skip
II. Gregor Mendel
-first idea of modern genetics
-Austrian (Czech Republic now) monk who discovered fundamental
principles of heredity
A. Background: (what we know now)
1. traits encoded in DNAchromosomes
2. geneseveral nucleotides that encode for a particular trait
3. allelevariation of gene
-gene for eye color
-alleles for blue, brown, green
4. locusparticular location of a gene on a chromosome
-always constant for particular genes
5. homologous chromosomes in a cell
a. each homologue contains all the same genes or same
traits at same loci-may be different alleles, but same genes
b. one homologue comes from each parent
c. 2 genes are a gene pair and together encode for one trait
d. ex: put drawing here
6. genotypeactual representation of alleles
-ex: Bb
7. phenotypeexpression of trait
-ex: brown eyes
-organisms can have same phenotype, but different
genotypes
8. pure individuals
a. homozygous dominant (BB)
-inheritance of dominant alleles
-dominant trait expressed
-ex: brown eyes
b. homozygous recessive (bb)
-2 recessive alleles
-recessive trait expressed
-ex: blue eyes
9. hybridheterozygous (Bb)
a. 2 different alleles are inherited
b. true dominant traitonly dominant allele is expressed
-ex: brown eyes
Basic Principles of Heredity Notes AP Biology Mrs. Laux
10. to indicate genotype in dominant/recessive situation:
Bcapitaldominant
blower caserecessive
B. Mendel’s peas
1. Mendel knew that breeding plants that were not pure-hybridsproduced plants of varying traits
2. used peas because:
a. easy to grow
b. may be self- or cross-fertilized to produce perfect flowers
(anther and stigma)
c. had many easily distinguishable traits-worked with 7
1. pod shapeinflated or wrinkled
2. flower positionaxial or terminal
3. seed coloryellow or green
4. pod colorgreen or yellow
5. seed shaperound or wrinkled
6.seed coat colorgrayish brown or white
7. stem lengthtall or short
3. Mendel began by breeding true (pure) pea plants in his crosses
a. self-fertilizedalways produced offspring identical to
themselves in those traits
4. crossed 2 pure plantsP generation
a. began by studying only one trait at a time
b. results in F1 (first filial) generationall offspring had only
one trait expressed
c. where did other trait go?
5. allowed F1 to self-pollinate
F2 (second filial) generation
Tall X short P
All tall
F1
3 tall:1 short F2
6. Mendel concluded that a “factor” for short flowers had not
disappeared, but was masked by the tall “factor”
7. also true in other 6 traits
-allelesalternate forms of gene (factor) that controls trait
-each trait2 alleles (one from each parent)
-nowhomologous chromosomes
-2 different allelesdominant over recessive
Mendel’s law of dominance
-2 alleles segregate during gamete formation
-sperm1 allele
-egg1 allele
-come together during fertilization to control
expression of trait in offspring
Basic Principles of Heredity Notes AP Biology Mrs. Laux
segregation (meiosis)
recombination (fertilization)
offspring
-pure individualcarry same allele
-all gametes
-hybrids50% of passing down one (allele) as
opposed to another
Mendel’s law of segregation
-allele pairs separate and recombine
Punnett Square
1. cross tall X short
P
TT X tt
Represent all possible
offspring that can result
100% hybrid
F1
Tt X tt
Monohybrid cross
cross between individuals that are hybrid for 1 trait
F2
25% (1/4) TT 1:2:1 probability of offspring with each genotype
50% (1/2) Tt
25% (1/4) tt
3:1 phenotype ratio
proves law of segregationtraits “blended”-would not see in F2 generation
Test cross
-short flowertt
-tall flowerTT
-to determine genotypetest cross
breeding of unknown with homozygous recessive
-ex: T_ X tt; if TT, all tall; if Tt, 50% tall, 50% short
Basic Principles of Heredity Notes AP Biology Mrs. Laux
-farmer conducts cross-any shorthybrid
-dominant X recessive does not equal short flowers; all tall
Dihybrid cross
-hybrid for 2 traits
-seed color/shape
P=YYRR X yyrr
F1=YyRr (dihybrid cross)
3:1 ratio? No-mixture
gametes
Yy
Rr
Y
y
R
r
YR
yR
Yr
Yr
4 possible gametes
Do a cross
Phenotypes
9 round, yellow (dominant)
3 wrinkled, yellow (dominant/recessive)
3 round, green (dominant/recessive)
1 wrinkled, green (recessive)
Led to Mendel’s Law of Independent Assortment
-each allele pair on separate chromosomes assort independently
Law of Probability
-Dice
-probability of rolling a 31/6
-rolling 3 on 2 die 1/6 X 1/6 = 1/36
1. Traits work the same way
a. segregation and recombination (at fertilization) are random events
b. therefore, knowing genotypes of parents, we can predict most likely
genotypes of offspring via laws of probability
c. probability 01
-0impossible
-1100%
1. probability of all possible outcomes must equal 1
-31/6
-not 35/6
_________
6/6 = 1
2. Rule of Multiplication (Product rule) “and statement”
a. probability that simultaneous events will occur is the product of their
individual probabilities
-ex: cross between hybrid tall flowered plants-probability of recessive
offspring? Must be tt
Monohybrid cross
egg: Ppsegregationp=1/2
sperm: Ppsegregationp=1/2
Recombination: ½ X ½ = ¼
Basic Principles of Heredity Notes AP Biology Mrs. Laux
b. Dihybrid cross
YyRr X YyRr
-probability that offspring YYRR
-segregation:
-egg (YyRr): Y=1/2, R=1/2 Both=1/4
-sperm (YyRr): Y=1/2, R=1/2 Both=1/4
-recombination:
-1/4 X ¼ = 1/16
3. Rule of addition (sum law) “or statement”
a. an event that can occur in more than one way is the sum of the
probabilities of each of those events
b. ex: when different genotypes result in the same phenotype
-ex: producing heterozygote from 2 hybrids
-egg: P (1/2) sperm: p (1/2) = ¼
-egg: p (1/2) sperm: P (1/2) = ¼
-¼+¼=½
4. easier way-memorize Punnett squares
5. solving more complex (trihybrid +) genetic probabilities:
-ex: probability of aabbcc from trihybrid cross of AaBbCc X AaBbCc
-each allele pair segregates independently
Aa X Aa
probability of aa is ¼
Bb X Bb
probability of bb is ¼
Cc X Cc
probability of cc is ¼
Total: ¼ X ¼ X ¼ = 1/64
____________________________________________
PpYyRr X Ppyyrr
-probability of recessive offspring?
Pp X Pp
¼
Yy X yy
½
Rr X rr
½
¼ X ½ X ½ = 1/16
_____________________________________________
PpYyRr X Ppyyrr
-probability that offspring will be recessive for at least 2 of 3 traits?
Could be:
ppyyRr
¼ X ½ X ½ = 1/16
ppYyrr
¼ X ½ X ½ = 1/16
Ppyyrr
½ X ½ X ½ = 2/16 = 1/8
PPyyrr
¼ X ½ X ½ = 1/16
Ppyyrr
¼ X ½ X ½ = 1/16
____________________________________________
6/16 = 3/8
Basic Principles of Heredity Notes AP Biology Mrs. Laux
-probability is not certain, only strongest possibility
-ex: Tt X Tt ¼ tt
-4 flowers-1 not necessarily short
-each time offspring is produced-1/4 chance it will be short
-larger amount of offspring (sample size), closer results will be to those
predicted
Dominance
1. not always completeinheritance characterized by one allele of
heterozygote that is completely expressed and other completely hidden
-ex: height
2. Dominance/recessiveness varies along a continuum from complete
dominance to codominance
Complete Dominance---------Incomplete Dominance----------Codominance
A is dominant
AA, Aa-same phenotype
A, B incomplete dominance
ABintermediate phenotype
(phenotype is intermediate
between A & B)
(no dominance)
AAAB both
alleles are
equally expressed
Incomplete Dominance
1. one allele is not completely dominant over the otherheterozygote has
a phenotype that is intermediate between 2 homozygotes
2. ex. Snapdragons (4 o’clocks)
RR (red)
WW (white)
F1
F2
-offspringpink traits appear
-proof that information for R and
blended, but each still maintains
W remains
information
3. Incompleteall alleles affect phenotype
Codominance
1. inheritance in which the heterozygous phenotype is a full expression of
both alleles
2. ex: roan cattle
CWCW-white
CRCR-red
F1
F2
Basic Principles of Heredity Notes AP Biology Mrs. Laux
All roan colorednot pink cows
Instead, have red and white hairs
Over whole body
show how separate alleles
remain in the heterozygote
Dominant/Recessive relationships
1. range: complete incomplete codominance
2. are a result of a mechanism that determines phenotypic expression, no ability
of one allele to subdue another at level of DNA
3. does not determine relative abundance of alleles in a population
-ex: polydactyly (1/400 births)
Pleiotrophy
1. ability of a single gene to have multiple phenotypic effects
2. one genemany phenotypes
3. ex: sickle cell anemia
a. abnormal hemoglobin molecule
b. sickle shape of RBC
c. prevalent in genetic disorders
4. sometimes unrelated phenotypes:
-ex: tigers and Siamese cats
a. gene controls fur pigmentation
b. also influences connection between eyes and cat’s brain
c. therefore, abnormal gene can cause both abnormal pigmentation and
cross-eyed condition
Epistasis
1. More than one geneone phenotype
2. Different genes can interact to control one phenotypic expression
3. a gene at one locus acts as an “on/off switch” for a gene at another locus
4. epistatic genea gene that has the ability to allow expression of suppression
of the other gene
5. occurs often in pigmentation
6. ex: micegene for pigment deposition (c) is epistatic to gene for melanin
(pigment) production
a. only if pigment is deposited in fur can color be seen
b. colors: brown/black: Bblack, bbrown
Pigment deposition: Cwill be deposited
cno deposition, therefore, no color
c. even though both genes affect the same trait (coat color), they are
inherited separately and will assort independently during gamete formation
d. ex: cross between black, heterozygous mice
BbCc X BbCc
Dihybrid cross
Result:
9 Black (B_C_), 3 Brown (bbC_), 4 Albino (__cc)
Basic Principles of Heredity Notes AP Biology Mrs. Laux
Polygenic Inheritance
1. Mendel’s traits all had only dominant and recessive alleles
-even heightT, t
2. not true of many traits
a. height in humans
-involves many genes that encode for the same trait
b. many gene pairsone phenotype
3. resulta continuous distribution of variations of the trait, not discrete either-or
differences
4. ex: skin pigmentation
a. controlled by at least 3 separately inherited genes
b. 3 genes (A, B, C) each contribute one “unit” of darkness to the
phenotype
c. each “add” to pigmentation
-AABBCC personvery dark
-aabbcc personvery light
-AaBbCc personintermediate shade
1900’sT.H. Morgan (Columbia)
1. provided convincing evidence that genes (Mendel’s factors) are located on
chromosomes
2. worked with fruit flies
Drosophila melanogaster
a. easily cultured
b. prolific breeders
c. short generation (2 weeks)
d. 4 pairs of chromosomes
1. 3 pairs autosomes (I, II, III)
2. one pair of sex chromosomes
3. Flieswild typemost common phenotype
mutantphenotype alternative to wild type that are caused by mutation
in a wild-type gene
4. bred flies
a. ~year, Morgan noticed a single male with white eyes (mutant)
-wild typered eyes
b. mated white eyed male with a red eyed femaleall red eyed offspring
(F1), therefore, recessive
1. crossed red eyed F1
2. result F2all females had red eyes, males-1/2 white, ½ red
5. Morgan deduced that eye color is linked to sex and gene for eye color is
located on X chromosome
-reasoning:
a. if eye color is located on X chromosome, then females (XX) carry 2
copies of genes, while males (XY) carry 1
b. for female to carry trait, mutant allele must be present on both Xs
Basic Principles of Heredity Notes AP Biology Mrs. Laux
c. malessince only one X, no wild type (dominant) allele to mask
recessive one; therefore, only need one copy of gene for expression of
trait
6. Sex-linked genes
a. genes located on sex chromosomes (usually X)
b. because of this, males tend to exhibit recessive sex linked traits more
often than females
c. ex: color blindness, hemophilia
7. Y-linked traits
a. those on Y chromosome
b. found only in males
c. ex: hairy ears, knuckles
8. Linked genes
a. genes on same chromosome tend to be linked in inheritance and do not
assort independently
b. move together during meiosis and fertilization
c. since independent assortment does not occur, dihybrid cross of 2 linked
genes does not result in a 9:3:3:1 ratio
Morgan looked at 2 mutations of Drosophila
1. body color and wing structure
a. wild type body color is gray (B)dominant
b. mutant (recessive)black (b)
c. wing structure: wild type (dominant) straight(V)
d. mutant (recessive) vestigial (small, underdeveloped, nonfunctional)(v)
2. performed test cross:
BbVv X bbvv
?
a. Mendel’s peas: expected test cross:
YyRr X yyrr
YyRr (1/2)(1/2)
yyrr (1/2)(1/2)
¼
¼
yyRr (1/2)(1/2)
Yyrr (1/2)(1/2)
¼
¼
phenotypic ratio:
1:1:1:1
b. therefore, if genes assorted independently, we would expect to see
equal numbers of flies with:
gray, straight wings (BbVv)
gray, vestigial wings (Bbvv)
black, straight wings (bbVv)
black, vestigial wings (bbvv)
c. what is genes are linked?
-since breeding gray, straight with black vestigial would expect:
Basic Principles of Heredity Notes AP Biology Mrs. Laux
-1/2 gray, straight, ½ black vestigial
d. actual results:
BbVv-965parental
Bbvv-185recombinant
bbVv-206recombinant
bbvv-944parental
ratio~41:41:9:9
e. because of overwhelming number of parental phenotypes, shows that
genes are linked, but
crossing over had occurred
resulted in recombinant phenotypes
f. 2 genes on chromosomeshigh % of crossing over-must be far from
each other on chromosome
g. to determine % of crossing over of 2 genes on a chromosome (or to
determine recombination frequency):
Recombination frequency = # of recombinants/total # of offspring X 100%
= 391/2300 = 17%
-shows that 17% of the time, crossing over does occur
exchange of parts between homologous chromosomes
forms recombinants with new allelic combinations
3. crossing over frequency can be used to determine relative loci of genes on a
chromosomes
a. higher the frequency, further apart the 2 genes are on the chromosome
-farther apart-more likely to cross over
b. because of this, recombination frequencies are used to give you a
picture of the arrangement of genes on a chromosome
1. can use %s to create a linkage mapportrayal of sequence of
genes on a chromosome
2. cytological maptrue gene positioning (much more advanced)
c. ex: fly with genotype BBVVAA (linked, epistatic gene)
(Aapterouswingless)
1. crossover frequency between B and V 18%
A and V 12%
B and A 6%
2. to make a linkage map, equate each % to a map unit
(arbitrary unitshows gene positioning)
B-V 18 mu (map units)
A-V 12 mu
B-A 6 mu